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3 years ago

2617 to the nearest thousand

Mathematics

2 answers:

Cloud [144]3 years ago

8 0

Answer:

2000

Step-by-step explanation:

because it is not over 5 if a number is over five then it is the next number is it is five then it's still the next number. Hope that helps

Sliva [168]3 years ago

7 0

Answer:

3000

Step-by-step explanation:

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Seventeen added to a number is sixty-five. What is the number?

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Answer:

problem: 17+x=65

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3 years ago

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Write 60 as the product of two factors

Ilya [14]

Use a factor tree to express 60 as a product of prime factors. So the prime factorization of 60 is 2 × 2 × 3 × 5, which can be written as 2 2 × 3 × 5.

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3 years ago

Can someone plz help me with #2, #3, #4, #14

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#2 is 5
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3 years ago

How do you do 3(4y-7)=15y+6

cestrela7 [59]

Answer:

y = 27/7

Step-by-step explanation:

Perform the indicated multiplication:

3(4y-7)=15y+6 becomes:

12y - 21 = 5y + 6

Combining like terms, we get:

7y = 27, so that y = 27/7

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4 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

6 0

3 years ago